Solve gives a hint 2 in r9c3, and I agree that a 9 here eventually creates a crash BUT I can't see any "logical" reason. I know this is probably a blind spot, but can't afford to waste more than an hour!

There's an X-Wing on the "2"s, in row 1 and row 5 (cols 1 & 9). So there's either a "2" at r1c1 and a "2" at r5c9, or else there's a "2" at r1c9 and a "2" at r5c1. Either way there cannot be a "2" at r8c1, leaving r9c3 as the only possible spot for a "2" in the bottom left 3x3 box.

There's also a swordfish on the "9"s (in columns 1, 6, and 8 -- rows 1, 7, & 8) which allows you to set a "2" at r1c9, but the X-Wing is certainly easier to spot.

So far as I'm aware, this X-Wing (or swordfish) cannot be avoided -- at least, I can't find a simpler clue. I think that's a first for the Daily Sudoku! dcb

Have you identified the cells occupied by the pair (68) in row 2? Once this is done, it should be straightforward to develop the puzzle to the position posted by George at the beginning of this thread and so get to grips with the comments which followed.

I might as well put my 2˘ in. After the initial elimination techniques, I noticed multiple X-Wing patterns, but they couldn't eliminate any candidates that I could see. There were also a couple of unique rectangles which I was unable to do anything with. I didn't look for swordfish yet, since it's such a hassle.

There was a "23" in r5c9 that looked promising as a starting cell for a forcing chain; the chain starting with the "2" left a "deadly pattern" and solving the cell with the "3" broke the puzzle wide open.

I hadn't yet moved to X-wings, Swordfish, etc. -- and I try not to. But the comments made by each contributor referenced the logic and not the name -- hence, I learned from those exchanges - and the solution makes sense.

This specific puzzle was referenced as 'very hard'. As it turned out, it was for me. But I had been sailing along on this one, while struggling with the May 1st puzzle. In reading the comments on the May 1st, the light suddenly dawned on me and I also understood the results to that one to.

Both puzzles were 'very hard' for me. And it really does not matter how they are classified. The important thing is that I can have the chance to ask for help when I hit the wall or offer help when needed.

WHAT THE H*** IS AN X-WING OR SWORDFISH. I AM NEW AT THIS SUDOKU STUFF AND I SOLVED THIS PUZZLE BY TRIAL AND ERROR. THE WAY I BROKE IT OPEN WAS THERE COULD EITHER BE A 2 OR 9 IN R9C5 AND I GUESSED 9 WHICH BROKE THE PUZZLE OPEN FOR ME.
PLEASE LET ME KNOW HOW TO SOLVE THIS PUZZLE USING LOGIC ONLY.
THANKS,

Quite a bit has already been written about the "advanced techniques" -- you can read some of that stuff from the links provided in another message.

Here, I'll explain the X-Wing one more time. I'm assuming you arrived at the position George Woods posted at the beginning of this thread.

Code:

*78 .5. .1*
5** .41 .**
1.6 .72 .35

85* 4*6 1**
*64 517 98*
7*1 **8 564

..7 185 2.6
*85 *6. 7.1
61* 7*4 358

I have marked every spot where a "2" can be legally placed with an asterisk. We reason as follows.

-- There are only two places to fit a "2" in row 1, at r1c1, or at r1c9.
-- Similarly, there are only two places to fit a "2" in row 5, at r5c1, or at r5c9.
-- If r1c1 = 2, then r5c9 = 2.
-- If r1c1 <> 2, then r1c9 = 2, and r5c1 = 2.
-- Either way, the "2" in column 1 must appear at either r1c1 or r5c1, and the "2" in column 9 must appear at either r1c9 or r5c9.
-- Therefore we cannot possibly have a "2" at r8c1, r2c9, or r4c9.

Now the grid looks like this:

Code:

*78 .5. .1*
5** .41 .*.
1.6 .72 .35

85* 4*6 1*.
*64 517 98*
7*1 **8 564

..7 185 2.6
.85 *6. 7.1
61* 7*4 358

We see that the only spot left for a "2" in the bottom left 3x3 box is at r9c3, and the "9" you found by trial and error [in r9c5] follows immediately.

A "swordfish" is similar to an X-Wing, but instead of involving 4 cells at the corners of a rectangle it involves 6, 7, 8, or 9 cells at the vertices of a 3 x 3 pattern (sort of like tic-tac-toe). The "complete" swordfish always has nine cells in it, but as many as three of those can be cells that are already resolved, leaving a skeleton that is still useful.

The logic is exactly the same as an X-Wing ... in the three columns and rows occupied by the swordfish cells, the target digit must lie within the pattern, and the target digit can be eliminated from cells outside that pattern.

The swordfish on "9"s appears as follows.

Code:

x78 *5x .1*
5.. .41 ...
1.6 .72 .35

85. 4.6 1..
.64 517 98.
7.1 ..8 564

x*7 185 2x6
x85 *6x 7x1
61. 7.4 358

Here I have marked the "corners" of the swordfish with the letter x, and I have used an asterisk to mark the four cells from which the digit "9" can be eliminated. In particular, this pattern allows one to set r1c9 = 2, and that breaks the rest of the puzzle wide open. dcb

I try my hand at the Sudoku problems on this website and after a lot of practice, have managed to solve all bar the occasional Very Hard puzzle. The one on 2nd May was all but impossible!! I thought I was doing well to get to the point of needing a Swordfish! I can understand Greg N's frustration. Thanks to David Bryant for his help. I am perplexed by these "in" terms for special moves, and Swordfish and X wings are completely new to me. This has put back my proper work by hours and hours.

There is a unique rectangle on <39> in R18 C46. R1C4 and R8C4 cannot be <9>.

There is a unique rectangle on <49> in R78 C18. R1C1 cannot be <2>.

There is a unique rectangle on <27> in R24 C89. R2C9 and R4C9 cannot be <2>.

Keith, I see a fair number of similar rectangles, but I can't do anything with them. For example, take the "39" rectangle. It's obvious that either r1c4=6 or r8c4=2, but I don't see enough because nothing seems locked. How do you deduce that both r1c4 and r8c4 cannot be "9"?

Thank you David, that's very helpful. I never thought in terms of what are the things that would leave the "deadly pattern." Also thanks to Keith who I know would have answered had you not beat him to the punch.

its a type 4 UR, because there are only the two 3's in column 4. So Davids chains can be written shorter:
r1c4=9 => r8c4=3 (=> r8c6=9 => r1c6=3 )
r8c4=9 => r1c4=3 (=> r1c6=9 => r8c6=3 )

Whenever you have a UR with a naked pair [39 above] in one row (column) and one of the numbers is strongly linked [the 3] in the other row (col), where there can be additional candidates, you can eliminate the second number from the UR corners in that row (col) [the 9].